How do you solve by completing the square 2x^2+7x-1=0?

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Answer 1 · 2015-04-01T02:07:16

#2x^2 + 7x - 1 = 0#

Remove the constant from the left-side by adding #1# to both sides
#2x^2+7x = 1#

Divide all terms by #2# to reduce the #x^2# coefficient to #1#
#x^2 + 7/2 x = 1/2#

If the first 2 terms of #(x+a)^2# evaluate as #x^2 + 7/2x#
then #a = 7/4#

Add (7/4)^2 to both sides of the equation
#x^2+7/2 x +(7/4)^2 = 1/2 + (49)/(16)#

Rewrite the left-side as a square (and simplify the right side)
#(x+7/4)^2 = (57)/(16)#

Take the square root of both sides (remember both plus and minus)
#x+7/4 = +-sqrt(57)/4#

#x = (-7 +sqrt(57))/4#
or
#x=(-7-sqrt(57))/4#